Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x - 5}{2x^2 - 30x + 100} \times \dfrac{-4x^2 + 400}{3x - 30} $
Solution: First factor out any common factors. $n = \dfrac{x - 5}{2(x^2 - 15x + 50)} \times \dfrac{-4(x^2 - 100)}{3(x - 10)} $ Then factor the quadratic expressions. $n = \dfrac {x - 5} {2(x - 5)(x - 10)} \times \dfrac {-4(x - 10)(x + 10)} {3(x - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {(x - 5) \times -4(x - 10)(x + 10) } { 2(x - 5)(x - 10) \times 3(x - 10)} $ $n = \dfrac {-4(x - 10)(x + 10)(x - 5)} {6(x - 5)(x - 10)(x - 10)} $ Notice that $(x - 5)$ and $(x - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-4(x - 10)(x + 10)\cancel{(x - 5)}} {6\cancel{(x - 5)}(x - 10)(x - 10)} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $n = \dfrac {-4\cancel{(x - 10)}(x + 10)\cancel{(x - 5)}} {6\cancel{(x - 5)}(x - 10)\cancel{(x - 10)}} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $n = \dfrac {-4(x + 10)} {6(x - 10)} $ $ n = \dfrac{-2(x + 10)}{3(x - 10)}; x \neq 5; x \neq 10 $